00:01
So in this problem, we have the reaction.
00:07
2i gas reversibly forms i2 gas.
00:13
And we're given the pressures at equilibrium of these two gases, and so we can find kp, and that's just the pressure of i2 over the pressure of i squared, and so that's 0 .21 over 0 .23 squared.
00:32
And so our kp is 3 .9698.
00:37
And so we have our kp and we'll come back to that.
00:42
But now we want to look at our q value when the volume halves to see which direction the equilibrium will shift or which direction the reaction will shift to reach equilibrium.
00:56
And so when our volume halves, our pressures double since they're inversely proportional.
01:03
So we just double each of our values from our k reaction to get a q value of 1 .98.
01:09
And so that means we're shifting to the left or towards reactants.
01:21
So now i'll go to a new page and rewrite the reaction to i -gas, reversibly forms i -2 -gas, and we'll set up our ice table.
01:35
So our pressures have doubled, so we'll use our new pressures, 0 .46 and 0 .42.
01:44
And then here we have minus 2x, and here we have plus x because of the reaction coefficients.
01:50
And so we have 0 .46 minus 2x and 0 .42 plus x as our new equilibrium pressures...